Question

Two perfectly conducting rails $1$ and $2$ lie perpendicular to a uniform magnetic field $B$(into the plane of paper as shown). Separation between the rails is $l$. Two bars $AB$ and $CD$ are placed perpendicular to the rails and move away from one another with velocity $v$. If the resistance of the bars are $R$ and $\frac{R}{2}$ respectively, the current in the bars is given by $\frac{xBlv}{3R}$. The value of $x^2$ is

Answer 1

Emf induced in each bar is $E=Blv$.
For $CD$, the end $D$ is positive and for $AB$, the end $B$ is positive, the equivalent electric circuit is as shown below.



Net current through the loop will be, $i=\frac{2E}{(R+\frac{R}{2})}=\frac{4E}{3R}=\frac{4Bvl}{3R}$
Comparing we get,
$\Rightarrow x=4\Rightarrow x^2=16$